Question

g I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.6 kg and radius r=0.37 m from rest at an initial height of h=6.7 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.7 m/s. What is the rotational inertia of my wheel?

Answer 1

Answer:

The rotational inertia of my wheel is  $I =1.083 \ kg \cdot m^2$

Explanation:

From the question we are told that

      The mass of the wheel is $m = 1.6 \ kg$

        The radius of the wheel is  $r = 0.37 \ m$

        The height is  $h = 6.7 m$

          The linear speed is  $v = 4.7 m/s$

According to the law of energy conservation

             $PE = KE + KE_R$

Where PE is the potential energy at the height h  which is mathematically represented as

             $PE = mgh$

While KE is  the kinetic energy at the bottom of height h

                 $KE = \frac{1}{2} mv^2$

Where  $KE_R$ is the rotational kinetic energy which is mathematically represented as

           $KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}$

Where  $I$ is the rotational inertia

       So  substituting this formula into the equation of energy conservation

         $mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}$

=>       $I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}$

substituting values

           $I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}$

             $I =1.083 \ kg \cdot m^2$